Chapter 12.2, Problem 44E

(a)

To determine

To Explain: the reasonable to use the exponential model to explain the relation between light intensity and depth.

Answer to Problem 44E

Scatter plot is not having strong curvature

Explanation of Solution

Given:

  

  

It is noticed that the given scatter plot is not having strong curvature and therefore it would be appropriate to utilize a linear model between the two variables of the scatter plot. Therefore it is appropriate to utilize a linear model between ln(intensity) and depth.

General linear model to expect ln(intensity) and time

  $\ln \left(\text{intensity}\right)=a+b\left(\text{depth}\right)$

Taking the exponential

  $\begin{array}{c}\text{intensity}=e^{\ln \left(\text{intensity}\right)}\\ =e^{a+b\left(\text{depth}\right)}\\ =e^a{e}^{b\left(\text{depth}\right)}\end{array}$

(b)

To determine

To find: the equation of the least-squares regression line.

Answer to Problem 44E

  $\ln \widehat{y}=6.78910-0.333021x$

Explanation of Solution

Given:

  

Calculation:

General equation for the least square regression line

  $\widehat{y}=b_0+b_{1}x$

The calculated of the constant $b_0$ is mention in the row “constant” and in the column “Coef” of the output of the computer

  $b_0$ =6.78910

The calculated of the slope $b_1$ is mention in the row “Depth” and in the column “Coef” of the output of the computer

  $b_1$ =-0.333021

Putting the value of $b_0$ and $b_1$

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \widehat{y}=6.78910-0.333021x\end{array}$

Where x shows the time and y shows the ln (count)

  $\ln \widehat{y}=6.78910-0.333021x$

Where $x$ is representing the depth and y is representing the intensity.

(c)

To determine

To find: the light intensity at depth of 12 meters.

Answer to Problem 44E

16.3275 lumens

Explanation of Solution

Given:

  

Calculation:

From the part (b)

  $\ln \widehat{y}=6.78910-0.333021x$

Putting the value of $x$

  $\begin{array}{c}\ln \widehat{y}=6.78910-0.333021x\\ \ln \widehat{y}=6.78910-0.333021\left(12\right)\\ =2.792848\end{array}$

Taking the exponential

  $\begin{array}{c}\widehat{y}=e^{\ln \widehat{y}}=e^{2.792848}\\ =16.3275\end{array}$

Therefore the expected that the intensity is 16.3275 lumens at depth of 12 metre